Confounded by King Arthur's knights

[Phoenixtears]

Homework Statement

King Arthur's knights fire a cannon from the top of the castle wall. The cannonball is fired at a speed of 46 m/s and at an angle of 25°. A cannonball that was accidentally dropped hits the moat below in 1.2 s.
(a) How far from the castle wall does the cannonball hit the ground?
m

(b) What is the ball's maximum height above the ground

Homework Equations

Vf= V0 +at
Deltax= V0t+ .5at^2
Vf^2= V0^2 +2ax
SOHCAHTOA

The Attempt at a Solution

I don't understand why I'm getting this wrong. I begin by separating it into two motions, horizontal and vertical. Using trigonometry I found that for the horizontal motion, the velocity was a constant 41.7 m/s. The Horizontal acceleration is 0, and the time is 1.2. I found these by using the listed equations above. 1.2 seconds is the time for both the vertical and the horizontal motions, by definition.

For the veritcle motion, the acceleration is a known -9.8, the time is constant between the two, the initial velocity is 19.4 (using the equations), the final velocity is -7.64 (using the equations), and the deltax is 16.2. Yet my answers are not working. Did I do something wrong mathematically. I've drawn the motion of the object, and I still don't understand.

I don't have the change horizontally becasue my number wasn't working right. For the second question, I made the final velocity zero horizontally (so that it was at the change of direction, the highest point) and then added that to my original 16.2. This was wrong as well. What am I doing incorrectly?

 

[mgb_phys]

Have you included the height of the wall? The falling cannon ball gives you the wall height.
You have the path of the fired ball correctly but then you have to all for the time for it to fall the extra wall_height.

Or perhaps it's because Arthur is a 6-8th century legend, at least 500years before cannon were general used in European battles?

 

[Phoenixtears]

I assumed that the vertical displacement was the height of the wall... I'll try to figure out if they're the same right now, and then try again.

Thank you!

PS- Physics problems never seem to be historically correct. :)

 

[Phoenixtears]

Hmm... I'm pretty sure that the vertical displacement <i>is</i> the height of the wall. Isn't the time merely given because the time is a constant? Besides, only two variables are given for the dropped cannon: 1.2 seconds and -9.8 m/s/s is the acceleration. Is it even possible to find velocity here? (To use d=vt)

 

[Phoenixtears]

Bah, nothing is working. I swear the answer to number two is 35.4, yet that isn't it. And for number one I use Deltax= V0t+ .5at^2 where acceleration is 0, and therefore elimiates the second part. Leaving V0t as the answer to the displacement. 41.7*1.2= 50.04, but that isn't working either.

 

[Defennder]

Note that you have to find the total time taken for the cannonball to reach its max height and then fall to the ground. These involve two separate calculations which you then have to add up.

Anyway it looks like the question is oddly structured since you would need to find (b) before you can do (a).